Proof regarding subspaces I have a problem (and the solution) about subspaces, I just don't understand it, any clarification would be appreciated.
Prove that if $S$ is a subspace of $ℝ^{1}$, then either $S={0}$ or $S = ℝ^{1}$.
The answer given:
If $S$ is a subspace of $ℝ$, then $0∈S$. It is easily verified that {0} is a subspace of $ℝ$. (I understand that $0∈S$ since a subspace is nonempty but I don't know this verification).
If $S$ contains other elements, i.e. the element $y≠0$ then $\alpha y∈S$ for any scalar $\alpha∈ℝ$, this implies $S=ℝ$ (I don't get why it does imply that)
Thanks in advance :)
 A: If $y\neq 0\in S$, then you know that for any $\alpha \in\mathbb R$, the vector $\alpha \cdot y$ is also in $S$ (otherwise, $S$ is not a subspace).
Now, take any $x\in\mathbb R$. Then, $$x=1\cdot x = \frac{y}{y}\cdot x=\frac{x}{y}\cdot y \in S$$
A: so the definition of a subspace is:
W is a subspace of V iff 
i) W is a linear space
ii) W is not empty
iii) $W \subset V$
You understand, that $\{0\}$ and $R$ are subset of W. So we just have to proof that there are no other spaces.
Assume W is a subspace of $R$ and $W$ is not $\{0\}$. Since removing the Element $0$, $\{0\}$ becomes empty. So there must be an element in W, which is not in $\{0\}$, e.g. there exists an x in W, $x\neq 0$. pick an y in $R$. If $y=0$ then $y \in W$ (W is a subset by assumption). if $y\neq 0$, then multiply x, with the scalar $a=y/x$. Since W is a subset $ax \in W$. But $ax=y$, so that $y \in W$. Since y was arbitrary, we have shown, that for all $y\in R$ $y\in W$, so that $R \subset W$. Because $W$ is a subspace of $R$, we know $W \subset R$. Combinig these, we get $W=R$.
This is a very detailed proof, I hope the details help:)
Cheers
A: Shortly said, $S$ is a subspace of $\mathbb{R}^n$ if two are satisfied: 


*

*$0 \in S$.

*for all $x,y \in S \implies \alpha x + \beta y \in S$.
Now consider two cases for $n = 1$. $S$ is either $\{ 0 \}$ or not. If not then $S = \{ 0 , \alpha x \}$ with $x \neq 0$ and $\alpha$ arbitary in $\mathbb{R}$, so $S = \mathbb{R}$.
